The graph of $y=ax^2+bx+c$ is given below, where $a$, $b$, and $c$ are integers.  Find $a-b+c$.

[asy]
size(150);

Label f;

f.p=fontsize(4);

xaxis(-3,3,Ticks(f, 1.0));

yaxis(-4,4,Ticks(f, 1.0));

real f(real x)

{

return x^2+2x-1;

}

draw(graph(f,-2.7,.7),linewidth(1),Arrows(6));
[/asy]
When $x=-1$, we have $y = a-b+c$.  The graph appears to pass through $(-1,-2)$.  Since $a$, $b$, and $c$ are integers, we know that $y$ is an integer when $x=-1$, so the graph does indeed pass through $(-1,-2)$.  Therefore, $y=-2$ when $x=-1$, so $a-b+c = \boxed{-2}$.